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第74页

信息发布者：

$-a$

$x + 1$

$ \begin{aligned} 解:原式&=\dfrac {b^2}{a^2}•\frac {3ac}{4b}×\frac {3a}{2b^2} \\ &=\dfrac {3bc}{4a}×\frac {3a}{2b^2} \\ &=\dfrac {9c}{8b} \\ \end{aligned}$

$ \begin{aligned}解:原式&=1-\dfrac {(x+1)(x-1)}{(x+1)^2}×\dfrac {x}{x-1} \\ &=\dfrac {(x+1)^2}{(x+1)^2}-\dfrac {x(x+1)}{(x+1)^2} \\ &=\dfrac {(x+1)(x+1-x)}{(x+1)^2} \\ &=\dfrac{1}{x+1} \\ \end{aligned}$

 解：原式$=\left(\frac{1}{x + 1}+\frac{x - 1}{x + 1}\right)\cdot\frac{x + 1}{x - 1}=\frac{x}{x + 1}\cdot\frac{x + 1}{x - 1}=\frac{x}{x - 1}。$

 $\because x$为整数，且$|x|\leq\sqrt{2}，$$\therefore x = \pm1$或$0。$

 $\because$当$x = \pm1$时，原分式无意义，

 $\therefore x = 0，$此时原式$=\frac{0}{0 - 1}=0。$